LeetCode Happy Number

原题链接在这里：https://leetcode.com/problems/happy-number/

题目：

Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

题解：

用HashSet 存储过往的和，若是遇见了相同的和，则表示出现了无限循环，没有happy number.

Note: 1.  Math.pow()的argument 和 return value 都是 double 型，返回时要注意cast.

遍历一个数字的每一位就用下面的代码：

while(n!=0){
    int digit = n%10;
    n = n/10;
}

AC Java:

public class Solution {
    public boolean isHappy(int n) {
        HashSet<Integer> hs = new HashSet<Integer>();
        
        while(!hs.contains(n)){
            if(n == 1){
                return true;
            }
            hs.add(n);
            
            int sum = 0;
            while(n != 0){
                int digit = n%10;
                n = n/10;
                sum += (int)Math.pow(digit,2);
            }
            n = sum;
        }
        return false;
    }
}

节省空间可以不需要HashSet<Integer> hs 来记录过往的digit square sum. 

可以类似Linked List Cycle用快慢指针找到loop的出口, 若出来时walker 和 runner 都是1就是happy number.

AC Java:

class Solution {
    public boolean isHappy(int n) {
        if(n <= 0){
            return false;
        }
        
        int walker = n;
        int runner = next(n);
        while(walker != runner){
            walker = next(walker);
            runner = next(next(runner));
        }
        
        return runner == 1;
    }
    
    private int next(int n){
        int res = 0;
        while(n > 0){
            int d = n % 10;
            res += d * d;
            n /= 10;
        }
        
        return res;
    }
}